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Chapter 1 : MATTER

1.1 Atoms and Molecules

1.2 Mole Concept

1.3 Stoichiometry

Chapter 1 : MATTER

1.1 Atoms and Molecules

1.2 Mole Concept

1.3 Stoichiometry

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Learning Outcomes

(a) Describe proton, electron and neutron in terms of mass and charge

(b) Define proton no, nucleon no and isotope

(c) Write isotopic notation

(d) Define relative atomic mass and relative molecular mass

(e) Calculate average atomic mass of an element

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At the end of the lesson, you should be ABLE to:

MATTER

Solutions

Mixtures

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5

What is ‘matter’?

Anything that occupies space/has volume and

has mass.

Example:

………………………………………………………..

Matter consists of 3 types of particles: atoms, molecules or ions.

Give an example of non-matter. Give your reason.

3 Types of Particles

ATOMSAn atom is the

smallest unit of a

chemical

element/compound.

Examples:

Na, Cu, Ne

MOLECULESA molecule consist of a

small number of atoms

joined together by bonds

Examples:

O2, CO2, H2O

IONSAn atom or molecule

with a net electric

charge due to the

loss or gain of one or

more electrons.

Examples:

Na+, NH4+,

O2-, NO3-

Example: Cuatom

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Atoms

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Diatomic Molecules

(e)

(n)

(p)

Atomic Structure

An atom consist of 3

subatomic particles:

Proton (p)

Neutron (n)

Electron (e) 9

Subatomic Particles

Subatomic

particle

Mass

(gram)

Charge

(coulomb)

Relative

charge

Electron

(e)9.1 x 10-28

-1.6 x 10-19

Proton

(p)1.67 x 10-24

+1.6 x 10-19

Neutron

(n)1.67 x 10-24 0

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• Proton number (Z) :

………………………………………………………..

• Nucleon number (A) :

…………………………………………………………

Proton number (Z) & Nucleon number (A)

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Element

symbol

(can be an

atom or an

ion!!!)

Proton

number

(p only)

Nucleon

number

(p + n)

An atom can be represented by an isotope

notation (atomic symbol) :

Isotope Notation

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Total charge

on the ion

proton number of

mercury,

Z = 80

Nucleon number of

mercury, A = 202

The number of

neutrons

= A – Z

= 202 – 80

= 122

* No of e = 80 -2 = 78

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Isotopes

..............................................................................................

..............................................................................................

............................................................................................

• Examples:

Hg200

80

2

1H U235

92

U238

92 3

1H

Hg202

80

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Ions (cation & anion)

Atom

(neutral)

Cation

(+ ion)

Anion

(- ion)Atom

loses e

Atom

accepts e

NaNa+

+ e Cl + e Cl-

Z = 11e = 11

Z = 11e = 10

Z = 17e = 17

Z = 17e = 18

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Exercise 1

Symbol Number of :

ChargeProton Neutron Electron

• Give the number of protons, neutrons, electrons and charge in each of the following species:

Cu63

29

217

8O

359

27Co

Hg200

80

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Exercise 2

Species

Number of :Notation

for nuclideProton Neutron Electron

A 2 2 2

B 1 2 0

C 1 1 1

D 7 7 10

• Write the appropriate notation for each of the following nuclide :

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Learning Outcomes

(a) Describe proton, electron and neutron in terms of mass and charge

(b) Define proton no, nucleon no and isotope

(c) Write isotopic notation

(d) Define relative atomic mass and relative molecular mass

(e) Calculate average atomic mass of an element

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At the end of the lesson, you should be ABLE to:

√√√

NEXT….

Relative Mass

Relative Atomic Mass (Ar)

Relative Molecular Mass (Mr)

Average mass of one

atom of an element

compared to 1/12

mass of one 12C atom

with the mass 12 g.

A mass of one molecule

of a compound compared

to 1/12 mass of one

atom

of 12C with the mass

12 g.

Ar =

Mr =

Both Ar and Mr are dimensionless; have no unit.

Relative Mass

Relative Atomic Mass (Ar)

Relative Molecular Mass (Mr)

Amu =atomic mass unit

Example 1

12-C of atom one of Mass 12

1

Y of atom one of mass Average

Determine the relative atomic mass of an element Y if the

ratio of the atomic mass of Y to carbon-12 atom is 0.45.

ANSWER :

Ar =

= 12 x 0.45

1

= 5.4

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The relative molecular mass of a compound is the SUMMATION of

the relative atomic masses of all atoms in a molecular formula.

NaOHAr of Na = 23.0

Ar of O = 16.0

Ar of H = 1.0

EXAMPLE 2:

Mr of NaOH= 23.0 + 16.0 + 1.0= 40

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Example 3 :

Calculate the relative molecular mass (Mr) of C5H5N

Answer :

Mr =

Learning Outcomes

(a) Describe proton, electron and neutron in terms of mass and charge

(b) Define proton no, nucleon no and isotope

(c) Write isotopic notation

(d) Define relative atomic mass and relative molecular mass

(e) Calculate average atomic mass of an element

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At the end of the lesson, you should be ABLE to:

√√√

NEXT….

√

Device used to measure Ar : Mass Spectrometer

Device used to measure Ar : Mass Spectrometer

Mass Spectrum

Mass Spectrum of Magnesium

• The mass spectrum of Mg shows that Mg consists of 3 isotopes : 24Mg, 25Mg and 26Mg.

• The height of each line is proportional to the abundance of each isotope.

• 24Mg is the most abundant of the three isotopes.

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8.19.1

24 25 26

Rela

tiv

e a

bu

nd

an

ce

(Q

)

m/e

(mass, amu)

% or

Ratio or

Fraction

m/e =

mass/charge

How to calculate the average atomic mass from mass spectrum?

Q = the relative / fractional / ratio / percentage abundance of isotopes of the element in the mixture

m = the isotopic mass of the element in unit amu.

The unit of average atomic mass is amu

Amu =atomic mass unit

Average

atomic

mass=

Example 1

Copper, Cu consists of two isotopes which is 69%

63Cu and 31% 65Cu. The isotopic mass of 63Cu and

65Cu are 62.9 a.m.u and 64.9 a.m.u respectively.

Calculate the average atomic mass of copper.

Average

atomic mass=

∑ Qi mi

∑ Qi

= (69 x 62.9 a.m.u) + (31 x 64.9 a.m.u)

69 + 31

= 63.52 a.m.u

69

31

62.9 64.9

Rela

tive

ab

un

dan

ce

(%)m/e (amu)

SOLUTION:

Example 2

a. What isotopes are present in Rb?

b. What is the percentage abundance of each isotope?

c. Calculate the relative atomic mass (Ar) of Rb.

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7

85 87

Re

lati

ve

ab

un

da

nc

e

m/e

(amu)

Given below is a mass spectrum of rubidium element, Rb.

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SOLUTION

18

7

85 87

Re

lati

ve

ab

un

da

nc

e

m/e

(amu)

(a) 85Rb and 87Rb

(b) % abundance 85Rb

= 18 X 100 = 72%

(18+7)

% abundance 85Rb

= 7 X 100 = 28%

(18+7)

85.56

amu x12.0012

1

amu 85.56Rb of A

amu 85.56

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)87x7()85x18(

Qi Rb of mass Average

r

iiMQ

CONTINUE SOLUTION

C.

The ratio of relative abundance of naturally occurring of chlorine

isotopes is as follow :

Based on the carbon-12 scale, the relative isotopic mass of

35Cl = 34.9689 and 37Cl = 36.9659. Calculate the Ar of

chlorine.

Example 3

3.127 37

35

Cl

Cl

SOLUTION

CONTINUE SOLUTION

Example 4

The relative atomic mass of Li and Li

are 6.01 and 7.02 respectively.

What is the percentage abundance of each

isotope if the relative atomic mass of Li is

6.94?

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SOLUTION

CONTINUE SOLUTION

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RecapWhat have you learnt today?